Mathematics HL/SL/Studies Help

[Markee]

@Derek so the answer key given by Mark is wrong?

@Mark no problem, one day you should help me in TOK or Chem

Does any one here knows how to use MS Excel by any chance?

For graphing? Yessh my physics teacher taught us how to plot stuffs for our lab reports anything I could do to help?

More on the TOK part and half on Chemistry. heh heh heh

And, well, I will be posting my IA Portfolio for Maths SL here, the whole thing, so you guys can see it and point out some tips and guides for me.

Mine is coincidentally on Infinite Summation

[dessskris]

@Derek so the answer key given by Mark is wrong?

@Mark no problem, one day you should help me in TOK or Chem

Does any one here knows how to use MS Excel by any chance?

For graphing? Yessh my physics teacher taught us how to plot stuffs for our lab reports anything I could do to help?

More on the TOK part and half on Chemistry. heh heh heh

And, well, I will be posting my IA Portfolio for Maths SL here, the whole thing, so you guys can see it and point out some tips and guides for me.

Mine is coincidentally on Infinite Summation

Not here, go to the infinite summtion thread for help on that IA. I can surely help you!

And don't post your IA in this forum. This task can still be used so people can later grab and plagiarise your IA.

[nametaken]

@Derek so the answer key given by Mark is wrong?

@Mark no problem, one day you should help me in TOK or Chem

Does any one here knows how to use MS Excel by any chance?

For graphing? Yessh my physics teacher taught us how to plot stuffs for our lab reports anything I could do to help?

The answer given by Mark is wrong, because when he was looking at the answers (at the back of the maths book), he accidentally misread it, and read the answer for another question. You're correct, Desy as the coefficient you've specified according to my calculations is correct. So, perception can often be misleading (a bit of TOK there for you!! )

[Markee]

@Derek so the answer key given by Mark is wrong?

@Mark no problem, one day you should help me in TOK or Chem

Does any one here knows how to use MS Excel by any chance?

For graphing? Yessh my physics teacher taught us how to plot stuffs for our lab reports anything I could do to help?

The answer given by Mark is wrong, because when he was looking at the answers (at the back of the maths book), he accidentally misread it, and read the answer for another question. You're correct, Desy as the coefficient you've specified according to my calculations is correct. So, perception can often be misleading (a bit of TOK there for you!! )

Ok bro, no need to point that out. Unless you want me to reveal you're secret identity. LOL!

But that perception of mine was taken over by emotion. And you know the rest

@Derek so the answer key given by Mark is wrong?

@Mark no problem, one day you should help me in TOK or Chem

Does any one here knows how to use MS Excel by any chance?

For graphing? Yessh my physics teacher taught us how to plot stuffs for our lab reports anything I could do to help?

More on the TOK part and half on Chemistry. heh heh heh

And, well, I will be posting my IA Portfolio for Maths SL here, the whole thing, so you guys can see it and point out some tips and guides for me.

Mine is coincidentally on Infinite Summation

Not here, go to the infinite summtion thread for help on that IA. I can surely help you!

And don't post your IA in this forum. This task can still be used so people can later grab and plagiarise your IA.

Sweetness! After the drafts I will heavily need your help

Really? Wow...Thanks!

[nametaken]

My brain has literally frozen....going into shut down mode, so I'd appreciate any help with a HL maths question, the question below is easy, but right now I just can't think:

The sum of an infinite geometric sequence is 13.5 and the sum of the first three terms is 13. Find the first term.

I've got the formula, S3= a(r^2-1)/r-1, which consequently = 13 and S infinity= a/r-1, which equals 13.5, but how do I go on from there?

Thanks for your help....whoever you are, wherever you are right now!!!!

[Markee]

Can anyone explain to me how Infinite Geometric Sequences works? We did this in class but our replacement teacher was just BAD He wasn't able to tame the class and wasn't able to explain it very well with examples...

I'd appreciate the assistance!

[nametaken]

Can anyone explain to me how Infinite Geometric Sequences works? We did this in class but our replacement teacher was just BAD He wasn't able to tame the class and wasn't able to explain it very well with examples...

I'd appreciate the assistance!

An infinite series is what I've explained above. It's a/r-1, that is it's the initial term divided by the common ratio (2nd term divided by first term) minus 1. An infinite series is infinite....it goes on and on and on and on and on and on and on and on and on and on and on and on and on and on and on and on................

[Markee]

Can anyone explain to me how Infinite Geometric Sequences works? We did this in class but our replacement teacher was just BAD He wasn't able to tame the class and wasn't able to explain it very well with examples...

I'd appreciate the assistance!

An infinite series is what I've explained above. It's a/r-1, that is it's the initial term divided by the common ratio (2nd term divided by first term) minus 1. An infinite series is infinite....it goes on and on and on and on and on and on and on and on and on and on and on and on and on and on and on and on................

LOL, haha. Thanks man

Could you be able to give me an example? If it's not asking too much ...

[nametaken]

Can anyone explain to me how Infinite Geometric Sequences works? We did this in class but our replacement teacher was just BAD He wasn't able to tame the class and wasn't able to explain it very well with examples...

I'd appreciate the assistance!

An infinite series is what I've explained above. It's a/r-1, that is it's the initial term divided by the common ratio (2nd term divided by first term) minus 1. An infinite series is infinite....it goes on and on and on and on and on and on and on and on and on and on and on and on and on and on and on and on................

LOL, haha. Thanks man

Could you be able to give me an example? If it's not asking too much ...

Well it depends on the question. If you're asked what an infinite geometric series is then the answer is that an infinite geometric series is an infinite sequence whose successive terms have a common ratio. That's probably confusing....but that's basically the definition. If you're stuck on a particular question, then ask away!

Edited February 8, 2011 by Derek

[dessskris]

@Mark sure PM me whenever you want me to read ok

btw the thing in your IA is not an infinite geometric summation. it's not geometric. any question regarding that should not be posted in this thread, should be in the infinite summation thread ok :)

---

My brain has literally frozen....going into shut down mode, so I'd appreciate any help with a HL maths question, the question below is easy, but right now I just can't think:

The sum of an infinite geometric sequence is 13.5 and the sum of the first three terms is 13. Find the first term.

I've got the formula, S3= a(r^2-1)/r-1, which consequently = 13 and S infinity= a/r-1, which equals 13.5, but how do I go on from there?

Thanks for your help....whoever you are, wherever you are right now!!!!

It's early in my place so I can't think. Assuming what you know now is correct,

13.5=a/(r-1)

a=13.5(r-1)

13=a(r^2-1)/(r-1)

13=13.5(r-1)(r^2-1)/(r-1)

13=13.5(r^2-1)

r^2=13/13.5 + 1

r=√(13/13.5 + 1)

a=13.5(r-1)

you calculate..as I said it's early here, I got to finish my HW so I don't have time to get this sorry

That's the final answer

[ILM]

Sorry for latness Derek. Anyway your question can be solved easily as Desy did. It is not a difficult question, but i think it has a mistake. Because r value will be bigger than 1 this mean you can not caculate the value of the infinite sequence, since it is diverging not converging. Please re-check the question, so we will be able to give a real numerical value. This what i had when i calculated the value of r, maybe i havea mistake, so please check that.

Edited February 9, 2011 by inm

[Sayyada Andani]

Helloo

I just wanted to ask if Maths Studies is accepted if I want to continue with studying pharmacy after IB.

For my HL subjects, I have Chem, Bio and Eco and at SL, I have English A1, Maths Studies SL and French Ab into

[dessskris]

Helloo

I just wanted to ask if Maths Studies is accepted if I want to continue with studying pharmacy after IB.

For my HL subjects, I have Chem, Bio and Eco and at SL, I have English A1, Maths Studies SL and French Ab into

Umm this is actually not the place :/ next time go to the university forum.

Nice you have HL Bio and Chem but Math Studies may make it hard for you to study pharmacy later. I would say at least Math Methods SL is required. A friend who took Math Methods SL and is now studying medicine is even having a hard time with the math, so I suspect you might be struggling very badly in pharmacy.

However it depends on where you want to go. Some unis may accept you but there are some others that will reject you.

[Julie]

My class did the intro to derivatives while I was sick... now we have a quiz tomorrow, and I dunno how to solve some stuff. Could someone solve one problem for me so I can get an idea of how to do them?

We were given the formula: y ' (x)= lim ((y(x+h)-y(x)) / h)

Then the problem from my math book is:

Use the algebraic/geometric method to find the slope of the tangent to:

y=x^2 at the point where x+3, i.e., (3,9)

Any help would be great

[Drake Glau]

My class did the intro to derivatives while I was sick... now we have a quiz tomorrow, and I dunno how to solve some stuff. Could someone solve one problem for me so I can get an idea of how to do them?

We were given the formula: y ' (x)= lim ((y(x+h)-y(x)) / h)

Then the problem from my math book is:

Use the algebraic/geometric method to find the slope of the tangent to:

y=x^2 at the point where x+3, i.e., (3,9)

Any help would be great

Well, using a short cut y'=2x so at x=3 the slope will be 6 But that's not how the book wants you to do it =/

y(x)=x² and y'(x) lim h->0=[y(x+h) - y(x)] / h

From here you can really just plug it all in to find your derivative eventually.

y'(x)=[(x+h)² - x²] / h

y'(x)=(x²+2xh+h² - x²) / h

x² cancel leaving (2xh+h²) / h

2xh / h=2x and h² / h = h leaving...

y'(x)=2x + h

Since it's a limit as h approaches zero you can really just kind of plug in zero for h now...

h=0 leaving you with just y'(x)=2x

Now that y'(x)=2x and you're looking for the slope at (3,9) you can plug in the 3 to get a slope of 6

Edited February 15, 2011 by Drake Glau

[genepeer]

y(x)=x² and y'(x) lim h->0=[y(x+h) - y(x)] / h

From here you can really just plug it all in to find your derivative eventually.

y'(x)=[(x+h)² - x²] / h

y'(x)=(x²+2xh+h² - x²) / h

y'(x)=2x + h

Since it's a limit as h approaches zero you can really just kind of plug in zero for h now...

That makes (x²) / h undefined and that end h=0 leaving you with just y'(x)=2x

Now that y'(x)=2x and you're looking for the slope at (3,9) you can plug in the 3 to get a slope of 6

corrected

[heyit'salison]

I am doing Mathematics HL and can't seem to understand Mathematical Induction!

Could you please help answering these:

1. Prove my mathematical induction:

a) 3*5 + 6*6 + 9*7 + 12*8 + ... + 3n(n+4) = [n(n+1)(2n+13)] / 2

b) 1^3 + 2^3 + 3^3 + 4^3 + ... + n^3 = [(n^2)(n+1)^2] / 4

and please show working out. i hope you understand the way i typed it!

thanks

[kiwi.at.heart]

Basically, the method for mathematical induction is:

1. Show true for n=1

2. Assume true for n=k

3. Show true for n=k+1

4. Statement showing that it is true

So...

a) 3*5 + 6*6 + 9*7 + 12*8 + ... + 3n(n+4) = [n(n+1)(2n+13)] / 2

Show true for n=1

3(1)((1)+4)=15 ((1)((1)+1)(2(1)+13))/2=15

LHS=RHS

Assume true for n=k

3*5+6*6+...+3k(k+4)=(k(k+1)(2k+13))/2

Show true for n=k+1

RHS=((k+1)((k+1)+1)(2(k+1)+13))/2

=((k+1)(k+2)(2k+15))/2

LHS=(k(k+1)(2k+13))/2 + k+1

From this point I was unable to find a way to get each side to equal each other (It might be because its been about 4 months since I have attempt mathematical induction or maybe you made a typo, I don't know), but usually you would get each side to be identical and hence showing LHS=RHS

Since true for n=k+1 and true for n+1, then it follows true for n=2, 3,...

b) 1^3 + 2^3 + 3^3 + 4^3 + ... + n^3 = [(n^2)(n+1)^2] / 4

If you follow the method from the first example, the next one is quite similar and you should be able to do it, but if you still have trouble just ask again

[dessskris]

I am doing Mathematics HL and can't seem to understand Mathematical Induction!

Could you please help answering these:

1. Prove my mathematical induction:

a) 3*5 + 6*6 + 9*7 + 12*8 + ... + 3n(n+4) = [n(n+1)(2n+13)] / 2

b) 1^3 + 2^3 + 3^3 + 4^3 + ... + n^3 = [(n^2)(n+1)^2] / 4

and please show working out. i hope you understand the way i typed it!

thanks

[dessskris]

Math HL

how do you integrate:

x arcsin(x)

with respect to x?

∫ x arcsin(x) dx = ??

I tried integrating by parts once and then by substitution twice but ended up with a very weird answer. I wonder if anybody knows how to integrate it like once or twice only instead of thrice. it would be great if they could also write it on a piece of paper and scan it because I guess the solution might be a bit long.. thank you!